Question

A bank wishes to estimate the mean credit card balance owed by its customers. The population standard deviation is estimated to be $300. If a 98 percent confidence interval is used and an interval of $78 is desired, how many customers should be sampled?A. 725B. 80C. 57D. 320

Answer 1

Answer: B. 80

Step-by-step explanation:

We know that the formula to find the sample size is given by :-

$n=(\dfrac{z^*\cdot\sigma}{E})^2$

, where $\sigma$ = population standard deviation.

E= margin of error

z*= Two -tailed critical z-value

Given : Confidence level = 98% =0.98

$\alpha=1-0.98=0.02$

Population standard deviation : $\sigma=300$

Also, from z-table for $\alpha/2=0.01$ (two tailed ), the critical will be = $z^*=2.326$

Then, the required sample size must be :

$n=(\dfrac{2.326\cdot300}{78})^2\\\\ n=(8.94615)^2\\\\ n=80.0336686391\approx80$  [To the nearest option]

Hence, the required sample size = 80

Hence, the correct option is option B. 80